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3 years ago

0.825167 rounded to the nearest ten-thousandth

Mathematics

1 answer:

antoniya [11.8K]3 years ago

3 0

0.825200
(correct me if I’m wrong)

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Mary invests £12000 in a saving account.

ladessa [460]

A=Value of investment.
A=12000(1+0.015)^2
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3 years ago

Ax+z=aw-y solve for a. show work please

dusya [7]

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3 years ago

The standard form of the equation of a parabola is y= x2 - 8x + 29.

fomenos

Answer:

Step-by-step explanation:

Put brackets around the first two tems.

y = (x^2 - 8x) + 29

Take 1/2 coefficient of the linear term -8. Square that result. Add it inside the brackets.

1/2 (- 8) = - 4

(- 4)^2 = 16

y = (x^2 - 8x + 16) + 29

Subtract 16 outside the brackets.

y = (x^2 - 8x + 16) + 29 - 16

Do the subtraction

y = (x^2 - 8x + 16) + 13

Represent what is inside the brackets as a square.

y = ( x - 4)^2 + 13

The answer is A

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3 years ago

Angie is organising a party for 84 adults and 42 children.

makvit [3.9K]

Add up the 84 and 42 to get 126, then divide by 6 to get the number of tables which is 21. And the number of seats is 126 because you would need a seat for every person. Hope this helps.

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3 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago